Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;
class Main
{
	public static void main (String[] args)
	{
		Scanner sc=new Scanner(System.in); 
		int i;	
		int n=sc.nextInt();
		double p[]=new double[n];
		for(i=0;i<n;i++)
			p[i]=sc.nextDouble();
		double dp[][]=new double[n][n+1];
		for(i=0;i<n;i++)
		{
			if(i==0)
				dp[i][0]=(1-p[i]);
			else
				dp[i][0]=dp[i-1][0]*(1-p[i]);
			for(int j=1;j<=i+1;j++)
			{
				if(i>0)
					dp[i][j]=dp[i-1][j-1]*p[i] + dp[i-1][j]*(1-p[i]);
				else
					dp[i][j]=p[i];
			}
		}
		double ans=0;
		for(i=(n+1)/2;i<=n;i++)
			ans+=dp[n-1][i];
		System.out.println(ans);
	}
}